9 Geometry

7/25/2019 9 Geometry

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M O D U L A R S Y S T E M

Class 9

GEOMETRY

?

w w w . z a m b a k . c o m


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Copyright Zambak Yaynclk veEitim Gereleri A..All rights reserved.

No part of this book may be reproduced,stored in a retrieval system or

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-112-4

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To

the

Teacher

Analytic Analysis of Lines and Circles is designed to provide students with the

analytic geometry background needed for further college-level geometry

courses. Analytic geometry can be defined as algebraic analysis applied to

geometrical concepts and figures, or the use of geometrical

concepts and figures to illustrate algebraic forms.

Analytic geometry has many applications in different

branches of science and makes it easier to solve a wide

variety of problems. The goal of this text is to help students

develop the skills necessary for solving analytic geometry

problems, and then help students apply these skills. By the

end of the book, students will have a good understanding

of the analytic approach to solving problems. In addition,

we have provided many systematic explanations throughout

the text that will help instructors to reach the goals that

they have set for their students. As always, we have taken

particular care to create a book that students can read,

understand, and enjoy, and that will help students gain

confidence in their ability to use analytic geometry.

To

the

Student

This book consists of two chapters, which cover analytical analysis of lines andcircles respectively. Each chapter begins with basic definitions, theorems, and

explanations which are necessary for understanding the subsequent chapter

material. In addition, each chapter is divided into subsections so that students

can follow the material easily.

Every subsection includes self-test heckYourself problem sections followed by basic

examples illustrating the relevant definition, theorem, rule, or property. Teachers

should encourage their students to solve Check Yourself problems themselves

because these problems are fundemental to understanding and learning the related

subjects or sections. The answers to most Check Yourself problems are given directly

after the problems, so that students have immediate feedback on their progress.

Answers to some Check Yourself problems are not included in the answer key, as they

are basic problems which are covered in detail in the preceding text or examples.


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Giving answers to such problems would effectively make the problems redundant,

so we have chosen to omit them, and leave students to find the basic answers

themselves.

At the end of every section there are exercises categorized according to the

structure and subject matter of the section. Exercises are graded in order,

from easy (at the beginning) to difficult (at the end).

Exercises which involve more ability and effort are

denoted by one or two stars. In addition, exercises which

deal with more than one subject are included in a

separate bank of mixed problems at the end of the

section. This organization allows the instructor to deal

with only part of a section if necessary and to easily determine which exercises

are appropriate to assign.

Every chapter ends with three important sections.

The hapterSummary is a list of important concepts and

formulas covered in the chapter that students can use

easily to get direct information whenever needed.

A oncept heck section contains questions about the

main concepts of the subjects

covered, especially about the definitions, theorems or

derived formulas.

Finally, a hapterReview Testsection consists of three tests, each with sixteen

carefully-selected problems. The first test covers

primitive and basic problems. The second and third tests

include more complex problems. These tests help

students assess their ability in understanding the

coverage of the chapter.

The answers to the exercises and the tests are given at the end of the book so

that students can compare their solution with the correct answer.

Each chapter also includes some subjects which are denoted as optional. These

subjects complement the topic and give some additional

information. However, completion of optional sections is

left to the discretion of the teacher, who can take into

account regional curriculum requirements.


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CHAPTER 1

SECTION 1:ANALYSIS OF VECTORS

GEOMETRICALLY

A. BASIC VECTOR CONCEPTS . . . . . . . . . .10

1. Directed Line Segment . . . . . . . . . . . . . . . . . . . .10

2. Definition of a Vector . . . . . . . . . . . . . . . . . . . . . .11

3. Equal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . .11

B. VECTOR OPERATIONS . . . . . . . . . . . . . .12

1. Addition of Vectors . . . . . . . . . . . . . . . . . . . . . . . .12

2. Subtraction of Vectors . . . . . . . . . . . . . . . . . . . . .16

3. Multiplication of a Vector by a Scalar . . . . . . . . . .17

C. PARALLEL VECTORS . . . . . . . . . . . . . . . .18

1. Parallel Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . .18

2. Non-Parallel Vectors . . . . . . . . . . . . . . . . . . . . . . .19

SECTION 2:ANALYSIS OF VECTORS

ANALYTICALLY

A. BASIC CONCEPTS OF VECTORS IN THE

ANALYTIC PLANE . . . . . . . . . . . . . . . . . . . .24

1. Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24

2. The Rectangular Coordinate System . . . . . . . . . .24

3. Position Vector . . . . . . . . . . . . . . . . . . . . . . . . . . .25

4. Components of a Vector . . . . . . . . . . . . . . . . . . .26

5. Equal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . .27

B. VECTOR OPERATIONS . . . . . . . . . . . . . .28

1. Addition of Vectors . . . . . . . . . . . . . . . . . . . . . . . .28

2. Subtraction of Vectors . . . . . . . . . . . . . . . . . . . . .29

3. Multiplication of a Vector by a Scalar . . . . . . . . . .30

4. Standard Base Vectors . . . . . . . . . . . . . . . . . . . . .31

C. PARALLEL VECTORS . . . . . . . . . . . . . . . .33

SECTION 3: THE DOT PRODUCT OF

TWO VECTORS

A. DOT PRODUCT . . . . . . . . . . . . . . . . . . .38

1. Properties of the Dot Product . . . . . . . . . . . . . . . .38

B. ANGLE BETWEEN TWO VECTORS . . . . .39

1. Angle Between Two Vectors . . . . . . . . . . . . . . . . .39

2. Perpendicular and Parallel Vectors . . . . . . . . . . .41

CHAPTER 2

SECTION 1: CONGRUENCE

A. THE CONCEPT OF CONGRUENCE . . . .56

1. Congruent Figures and Polygons . . . . . . . . . . . .56

2. Congruent Triangles . . . . . . . . . . . . . . . . . . . . . . .57

B. THE TRIANGLE ANGLE BISECTORTHEOREM . . . . . . . . . . . . . . . . . . . . . . . . .64

SECTION 2: THE CONCEPT OF

SIMILARITY

INTRODUCTION TO SIMILARITY . . . . . . . .67

1. Similar Figures . . . . . . . . . . . . . . . . . . . . . . . . . . .67

2. Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . .67

SECTION 3: THE ANGLE - ANGLE

SIMILARITY POSTULATETHE ANGLE-ANGLE (AA) SIMILARITYPOSTULATE . . . . . . . . . . . . . . . . . . . . . . . . . .73

SECTION 4: WORKING WITH SIMILARITY

TRIANGLES

A. THE SIDE-ANGLE-SIDE (SAS)SIMILARITY THEOREM . . . . . . . . . . . . . . . .83

B. THE SIDE-SIDE-SIDE (SSS)SIMILARITY THEOREM . . . . . . . . . . . . . . . .85

C. THE TRIANGLE PROPORTIONALITYTHEOREM AND THALES THEOREM . . . . . .90

1. The Triangle Proportionality Theorem . . . . . . . . .90

2. Thales Theorem of Parallel Lines . . . . . . . . . . . .93

D. FURTHER APPLICATIONS . . . . . . . . . . . .95

1. Menelaus Theorem . . . . . . . . . . . . . . . . . . . . . . .952. Cevas Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .96

SECTION 5: FURTHER STUDIES

A. EUCLIDEAN RELATIONS . . . . . . . . . . .103

B. MEDIAN RELATIONS . . . . . . . . . . . . . .106


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SECTION 3: SOLIDS WITH CURVED

SURFACES

A. Some Important Polyhedrons . . . . . . . .300

1. Prisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .300

SECTION 4:VOLUMES OF SOLIDS

A. Circular Cylinder . . . . . . . . . . . . . . . . .320

B. Areas of Cones . . . . . . . . . . . . . . . . . .323

C. Spheres . . . . . . . . . . . . . . . . . . . . . . .325

1. Fundamental Definitions . . . . . . . . . . . . . . . . . . .325

SECTION 5: SECTIONS AND

COMBINATIONS OF SOLIDS

1. Volume of a Right Prism . . . . . . . . . . . . . . . . . . .330

2. Volume of a Pyramid . . . . . . . . . . . . . . . . . . . . .331

Volumes of Cones . . . . . . . . . . . . . . . . . .336


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Geometry 9

Some of the quantities we measure in our daily lives are completely determined by their

magnitudes, for example, length, mass, area, temperature, and energy. When we speak of a

length of 3 cm or an area of 5 cm2, we only need one number to describe each of these

quantities. We call such quantities scalarquantities.

On the other hand, to describe a force, we need to record its direction as well as its size. For

example, to describe the velocity of a moving object, we must specify both the speed and the

direction of travel. Quantities such as displacement, velocity, acceleration, and other forces

that have magnitude as well as direction are called vectorquantities. We usually show a

vector quantity as an arrow that points in the direction of the action, with length that shows

the magnitude of the action in terms of a suitable unit. The way to represent such quantities

mathematically is through the use of vectors.

1. Directed Line Segment

When we move from Antalya to Berlin

by bus, we have two quantities: the

direction from Antalya to Berlin, and

the length of the displacement between

these cities.

We can sketch a line segment AB as shown in the figure with starting

pointA and finishing point B to represent the movement from Antalya to

Berlin. The line segmentAB with an arrow has direction and length. The

arrow head specifies the direction, and the length of the arrow specifies

the magnitude, at a suitable scale.A and B are the endpoints of the segment.

PointA is called the initialpoint and point B is called the terminalpoint

of the line segment. The resulting segmentAB is called a directedlinesegment.

Definition directedlinesegment

A line segment with direction is called a directedlinesegment.

We write

B to denote a directed line segment from pointA to point B.

Directed line segments are used in daily life. For example, some

traffic signs for drivers use directed line segments.

In technology we also use directed line segments.

A. BASIC VECTOR CONCEPTS


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Analysis of Vectors Geometrically

PointsM, N, P and K on line d are given. Write all the

directed line segments with endpointsM, N, P, or K.

The directed line segments with endpoints M, N, P, or K are

MN,

MP,

MK,

NP,

NK,NM,

PK,

PN,

PM,

KP,

KN, and

KM.

Notice that

MN is not the same as

NM, and

MP is not the same as

PM. This is because the

line segments have direction. Pairs such as

MN and

NM have the same magnitude but

opposite direction.

Solution

EXAMPLE 1

Definition vector

A directed line segment in the plane is called a vector.

The length of the directed line segment is the length of the

vector.

The direction of the directed line segment is the direction of

the vector.

We write

B to mean a vector with initial pointA and terminal point B. Alternatively, we can

name a vector with a lower-case letter such as u or p.

3. Equal Vectors

Definition equalvectors

Two vectors that have the same direction and length arecalled equalvectors. We show that two vectors

u and

v are

equal by writing u =

v.

In the figure, D, E, and F are the midpoints ofAB,AC and

BC respectively, and DE || BC, EF ||AB, DF ||AC.

Name all the equal vectors.

In triangle ABC,|DE| = |BF| = |FC|

|EF| = |AD| = |DB|

|DF| = |AE| = |EC|.

So

DE =

BF =

FC

EF =

AD =

DB

DF =

AE =

EC

ED =

FB =

CF

FE =

DA =

BD

FD =

EA =

CE.

Solution

EXAMPLE 2

2. Definition of a Vector

and

For example, consider a line segment

AB with length 2 cm.

We can say the length of vector

AB is 2 cm, and write |

AB| = 2 cm.


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2

Geometry 9

Ch e ck Yo u r se l f

ABCD is a parallelogram in the figure.

1 How many pairs of equivalent directed line segments are there?

2 How many pairs of equal vectors are there?

A n sw e r s

1 4 2 4

1. Addition of Vectors

Let

PQ and

QRbe two vectors in a plane.

PQ +

QR denotes the sum of the vectors

PQ and

QR. There are two ways to find the sum of two or more vectors.

a. The Polygon Method

Imagine we want to add n vectors together. Using the polygon method, we draw the first

vector. Then we place the initial point of the second vector at the terminal point of the first

vector, the initial point of the third vector at the terminal point of the second vector, and so

on until we place the initial point of the nth vector at the terminal point of the (n 1)th

vector. The sum is the vector whose initial point is the initial point of the first vector and

whose terminal point is the terminal point of the last vector.

Let us look at an example.Let

AB and

CDbe two vectors in a plane, as in the

diagram. We place the initial point of

AB at the

terminal point of

CD to make

DE (

AB =

DE).

Using the polygon method,

CD +

AB =

CD +

DE =

CE.

B. VECTOR OPERATIONS

Definition oppositevectors

Two vectors are called opposite vectors if and only if their

magnitudes (lengths) are the same but their directions are opposite.

Definition zerovector

A vector whose initial and terminal points are the same is called a zerovector.

We write a zero vector as0.

A zero vector has no direction and no size.

For example, in the figure,

AB and

BA are opposite vectors.

CD and

DC are also opposite vectors. We can write

AB =

BA and

CD =

DC.


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3


b. The Parallelogram Method

To add one vector to another using the parallelogram method, we draw the first vector, and

then we draw the second vector with its initial point at the initial point of the first vector. We

make a parallelogram by drawing two additional sides, each passing through the terminal

point of one of the vectors and parallel to the other vector. We find the sum by drawing a

vector along the diagonal from the common initial point to the intersection of the two lines.

Look at the example of addingu and

v using the parallelogram method:

Now look at an example of adding more than two vectors using the polygon method.

As shown in the figure,u +

v +

w +

x =

AE.

Findu +

v +

w in the figure on the right.EXAMPLE 3


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4

Geometry 9

Let us choose A as a fixed point. We can use the polygon method or the parallelogram

method to add the given vectorsu +

v +

w .

Solution

The velocity of a boat is 25 m/min north and the velocity of a river current is 3 m/min east.

Draw a scale diagram to show the velocities as vectors and find the sum.

First we choose a starting point A and

write

AN = velocity of the boat due north.

AE = velocity of the current due east.

AN and

AE are perpendicular, and

AK is

the sum of

AN and

AE:

|

AK| =

This is the sum of the vectors.

2 225 + 3 = 634.

Solution

EXAMPLE 4

c. Properties of Vector Addition

Letu ,v, and

w be three vectors in a plane P.

1 The sum of any two vectors in P is

also a vector in P (closure property).

2 The sum of any two vectors in P is

commutative (commutative property).


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5


4 The sum of the zero vector and a vector

in P is the vector itself (identity

element).

3 The sum of any three vectors in P is associative (associative property).

u +(

v +

w ) = (

u +

v ) +

w

5 The additive inverse of any vectoru is

u:u + (

u) =

0 (additive inverse).

In a triangleABC, P is the midpoint of

AB. Express

CP in terms of

CA and

CB.

CP = CA + AP

+

CP =

CB +

BP

2

CP =

CA +

CB +

AP +

BP

CP = (

CA +

CB)1

2

Solution

EXAMPLE

0


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6

Geometry 9


1 Find the following using the vectors in the figure.

a

v u b u + w c w + v u

2 In a triangleABC, D [BC] and |BD| = 2 |DC|.

Express

AD in terms of

AB and

AC.

A n sw e r s

1 use the polygon method 2

AD =

AC +

AB1

3

2

3

2. Subtraction of Vectors

Since subtraction is the inverse of addition, we can find the difference of two vectorsu and

v by adding the vectors

u and

v (opposite of

v) using either the parallelogram method

(u v = u + (v )) or the polygon method.

In a triangleABC, G is the centroid. Find

GA +

GB +

GC.

Let us label a point G on the extension of CG which

satisfies |CG| = |GG|. Since G is the centroid of

ABC, |CG| = 2|GK|. Therefore |GG| = 2|GK|,which means that K is the midpoint of GG. We con-

clude thatAGBG is a parallelogram because K is the

midpoint of both diagonalsAB and GG. So we have

AG =

GB which gives us

GA +

GB =

GG.

On the other hand, we have

CG =

GG =

GC. Using

this result in

GA +

GB =

GG, we get

GA +

GB =

GC which gives us

GA +

GB +

CG =

0.

Solution


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7


3. Multiplication of a Vector by a Scalar

Multiplying a vector by a scalar makes the vector longer or shorter depending on the value of

the scalar. If the scalar is greater than 1 or less than 1, multiplying makes a longer vector. If

the scalar is between 1 and 1 and non-zero, it makes a shorter vector.If the scalar is positive, multiplying does not change the direction.

If the scalar is negative, multiplying will make the vectors direction opposite.

For a real number a and a vectoru,

1 if a > 0 then vector au has the same direction as

u and the length |a

u| = a|

u|.

2 if a < 0 then vector au has the opposite direction to

u and the length |a

u| = |a||

u|.

3 if a = 0 then a

u =

0.

Using

AB as shown in the figure, draw vector

diagrams to show 2

AB, 4

AB, and

AB.1

2

Since 2 and are positive, 2

AB and

AB have the

same direction as

AB. However, 2

AB is twice as long

as

AB and

AB is half as long.

On the other hand, 4

AB has opposite direction to

AB (since 4 is a negative scalar) and it is four times

as long as

AB.

1

2

1

2

1

2Solution

EXAMPLE 5

a. Properties of the Multiplication of a Vector by a Scalar

For any vectorsu,v, and

w and real numbers a and b, the following properties are satisfied.

1 a

u is a vector in the plane2 (ab)

u = a(b

u )

3 (a + b)u = a

u + b

u

4 a(u +

v ) = a

u + b

v

5 1u =

u

6 a

0 =

0


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8

Geometry 9


1 Multiply the vector

u by the scalars 2, 3, 0.5 andand draw a vector diagram to show them.

2 PointsA, B, C, andM are on the same line.M is between

A and C.

AB = 2

AC. Express the vector

MC in terms of

the vectors

MA and

MB.

A n sw e r s

2

MC = (

MA +

MB)1

2

1

3

1. Parallel Vectors

C. PARALLEL VECTORS

Definition parallelvectors

Leta and

bbe two vectors.

a and

b are called parallelvectors if and only if

a = k

b where

k 0 and k . We write

a

||

b to show that two vectors are parallel.

For example, in the diagram, |a| = 2 cm,

|b| = 1 cm and |

c| = 4 cm.

We can express vectora as

a =

c and

a = 2

b.

Therefore the vectorsa,b, and

c are parallel, i.e.

a||b||c.

1

2

PointsA, B, C, andM are on the same line.M is between C and B.

AB = 3

AC. Express the

vector

MC in terms of vectors

MA and

MB.

AB = 3

AC so

CB = 2

AC (1)

MA +

AC =

MC (2)

CM +

MB = 2

AC (3)

AC =

MC +

MB (4)

MA

MC +

MB =

MCby (2) and (4).

MA +

MB =

MC +

MC

MA +

MB =

MC

So

MC =

MA +

MB.1

3

2

3

3

2

1

2

12

12

1

2

1

2

1

2

1

2

Solution

EXAMPLE 6


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9


In a triangleABC, D and E are the midpoints of sidesAB andAC respectively.

Show that

DE ||

BC.

EXAMPLE 7

BA +

AC =

BC and

DA +

AE =

DEby the addition of vectors.

DA =

BA,

AE =

AC

DE =

BA +

AC = (

BA +

AC) =

BC

Now

DE =

BC, so

DE ||

BCby the definition of parallel vectors.1

2

1

2

1

2

1

2

1

2

1

2

1

2

Solution

BA +

AC =

BC (1) and

EA +

AF =

EF (2).

EA =

BA

+

AF =

AC

EA +

AF = (

BA +

AC)

EF =

BCby (1) and (2). Therefore,

EF ||

BC.1

2

1

2

1

2

1

2

Solution

EXAMPLE8

In a quadrilateral ABCD, points E and F are the midpoints of side AB and diagonal AC,respectively. Show that EF || BC.

2. Non-Parallel Vectors

By the definiton of parallel vectors we can conclude that ifa and

b are non-zero,

non-parallel vectors, then h

a = k

b when h =k = 0. Look at the proof:Suppose that h k 0,

a =

b.

Thena||b. This is a contradiction, since

a and

b are non-parallel. As a result, h = k = 0.

k

h

EXAMPLE 9 Prove that the diagonals of a parallelogram intersect at their midpoints by using vectors.

Look at the diagram. Let

AB =

a and

BC =

b, so

AC =a +

b

DB =a b

AE =m(a +

b)

EB =n(a b )

Solution


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2

Geometry 9


1 Name all the pairs of parallel

vectors in the figure.

2 Add the vector pairs

u1 and

u5,u2 and

u6, and

u3 and

u7.

3 Find the additive inverse ofu7

andu1.

4 Subtractu8 from

u4.

5 In a quadrilateralABCD, P and

K are the midpoints of sidesAB

and DC respectively. Express

PKin terms of

DA and

CB.

A n sw e r s

1 look at the directions and lengths 2 use the polygon method 3 use the polygon method

4 use the polygon method 5

PK = (

DA +

CB)1

2

Let us draw the triangleABC as in the figure.

If D, E and F are midpoints then

AD = (

AB +

AC) by the result of Example 5.

BF = (

BA +

BC)

+

CE = (

CA +

CB)

AD +

BF +

CE = (

AB +

BA +

AC +

CA +

BC +

CB )

Therefore,

AD +

BF +

CE =0.

1

2

1

2

1

2

1

2

Solution

EXAMPLE 10 In a triangleABC, D, E and F are the midpoints of sides BC, BA andAC respectively. Find thesum

AD +

BF +

CE.

0

0

0

AE +

EB =m(a +

b ) +n(

a b )

a =m(

a +

b) +n(

a b)

(m +n 1)a =

b(n m).

Since

a and

b are non-zero and non-parallel, we have (m +n 1) = (n m) = 0. Therefore

m +n = 1 andm =n, and som =n = .1

2


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2


Project:Describe some other areas in which we use directed line segments

(for example: flowcharts, keyboards, ...).

A. Basic Vector Concepts

1. Draw any two parallel directed line segments withthe same length but opposite direction. Express

one of the line segments in terms of the other.

EXERCISES 1

2. How many equal directed line segments can we

find on two parallel lines?

3. Make a scale diagram showing the vectors in each

statement and find their sum.

a A 6 km trip east is followed by a 3 km trip

southeast.

b The velocity of a swimmer is 5 m/min west

and the velocity of a river current is 2 m/min

north.

4. Using the vectors given on the right,sketch the following vectors.

a

u +

v b

w (

u +

v)

c w

v

u d

u + (

w +

v)

e

u 3

v + 2

w

B. Vector Operations

5. In a plane, [AB] is given. Point K is the midpoint

of [AB] and point O is any point in the same

plane. Express

OK in terms of

OA and

OB.

6. In a triangleABC, points D and E lie on [BC] and

|BD| = |DE| = |EC|. Express the vector

AD +

AE

in terms of

AB and

AC.


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22


12. In a six-sided polygonABCDEF,

AB =

ED,

BC =

FE,

and

CD =

AF. Show that FBCE is a parallelogram.

In a parallelogram ABCD, K is the midpoint of

side DC, and the intersection point of diagonals

AC and BK is T. Show that

AC = 3

TC.

18.

In a quadrilateralABCD, E and F are the midpoints

of the diagonalsAC and BD respectively.

Show that

AB +

AD +

CB +

CD = 4

EF.

15.

14. In a triangle ABC, |BD| = |DE| = |EC|, and

E, D [BC]. If |

AD+

AE| = 9 cm, find |

AB +

AC|.

Show that the centroid of a triangle divides a

median in the ratio 1:2 using vectors.

17.

13. A trapezoid is a four-sided figure with only two

parallel sides. A line segment which joins the

midpoints of the non-parallel sides is called the

median of the trapezoid. Prove that the median of

a trapezoid is parallel to the two parallel sides,and has magnitude equal to half of their sum.

In the figure,

Tis the midpoint of BC,

ABC is a triangle,

2|AK| = |KB|, and

2|AM| = |MC|.

Use vectors to show that

|

AL| = |

LT|.1

2

16.

11. ABCD is a quadrilateral and M, N, P, Q are the

midpoints of AB, BC, CD, and DA respectively.

Show thatMNPQ is a parallelogram.

C. Parallel Vectors

10. In a triangleABC, G is the point of intersection of the

medians and P is the midpoint of

BG. Show thatPA +

PC = 4

PG.

Consider any two points A and B in a plane.For

any point P in the same plane, the symmetry of

point P with respect to point A is Q and the

symmetry of point Q with respect to point B is

point R. Show that

PR is always 2

AB. (Hint: letA,

B, and Cbe collinear. If |AB| = |BC| then A is

the symmetry of C with respect to B.)

9.

8. Point O is in the plane of a triangle ABC. Point G

is the centroid of triangle ABC. Show that

OA +

OB +

OC = 3

OG.

7. Show that (

AB +

BC) +

CD =

AB + (

BC +

CD)

by using the parallelogram method.


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24

Geometry 9

We have studied vectors geometrically. Now let us look at a method for describing vectorsanalytically.

We will begin this section by looking at some important axioms.

1. Axioms1 For each pair of points P and Q there exists a unique vector

v such that

PQ =v.

2 For each point P and vectorv, there is a

unique point Q such thatv =

PQ.

A. BASIC CONCEPTS OF VECTORS IN THE ANALYTIC

PLANE

Con c l u s i o n

1 Two points in a plane determine two opposite vectors.

2 In a plane, if one point is fixed as an initial point then all the other points in the plane

can be chosen as the terminal point of any vector.

2. The Rectangular Coordinate System

The rectangularcoordinatesystem is formed by

two perpendicular intersecting number lines, as

shown in the diagram opposite.

1 The horizontal number line is called the x-axis.

2 The vertical number line is called the y-axis.

The origin is the point of intersection. At this

intersection, both number lines are 0. The

rectangular coordinate system is split into four

quadrants, which are marked in the diagram

with roman numerals.

Each point in the coordinate system is associated with a pairofrealnumbers. In an x, y

system, the x-coordinate always comes first and the y-coordinate always comes second in the

pair (x, y). The first coordinate is called the abscissa of the point and the second coordinate

is called the ordinate of the point.


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Analysis of Vectors Analytically

A(2, 3) lies in quadrant I.

B(1, 2) lies in quadrant II.

C(3, 4) lies in quadrant III.

D(2, 0) lies on the x-axis.

E(0, 5) lies on the y-axis.

Solution

EXAMPLE 11 Plot each pair of coordinates and name the quadrant or axis in which the point lies.A(2, 3), B(1, 2), C(3, 4), D(2, 0), E(0, 5)

Ch e c k Y o u r s e l f

Plot the pairs of real numbers and name the quadrant or axis in which the point lies.

A(3, 2), B(2, 1), C(4, 3), D(0, 2), E(5, 0)

From this definition we can conclude the following:

1 For every vector in the plane there exists a position

vector

OP which is determined by a pair (x, y), that is,

OP = (x, y).

2 If

OP =

AB, then OPBA is a parallelogram.Look at the diagram. We can calculate that x = x2 x1 and

y = y2 y1. Therefore the vector

AB determined by the

pointsA(x1, y1) and B(x2, y2) has position vector

OP =

AB =

OB

OA = (x, y) = (x2 x1, y2 y1).

OP =

KL =

OL

OK = (4 2,5 1)

OP = (2, 4)

Solution

EXAMPLE

12Find the position vector of

KL with endpoints

K(2, 1) and L(4, 5).

3. Position Vector

Definition positionvector

Avector

OP whose initial point is at the origin of the rectangular coordinate plane and which

is parallel to a vector

AB is called the positionvector of

AB in the plane. In other words, ifOP is the position vector of

AB, then

OP ||

AB, |

OP| = |

AB|, and

OP =

AB.


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26

Geometry 9

4. Components of a VectorWe have seen how to describe a vector in the plane with

reference to its unique position vector.

We can also express a vector in the plane as the sum of

two vectors, one of which is parallel to the x-axis and the

other parallel to the y-axis. These two vectors are called

the components of the original vector. The component

parallel to the x-axis is called the horizontalcomponent

of the vector and the component parallel to y-axis is

called the verticalcomponent. Expressing a vector as the

sum of its components like this is called resolving the vector. For example, in the figure

opposite, the vector

u is the sum of the two components

ux and

uy.We can represent the vector

u as an ordered pair of real numbers:

u = (u1, u2) or

u = ,

where u1 is the horizontal scalar component ofu, and

u2 is the vertical scalar component ofu.

Now,ux = (u1, 0) and

uy = (0, u2).

Look at the diagram opposite. By applying thePythagorean theorem for triangle PQR in the figure, we

can see that the length of the vectoru = (u1, u2) is

|u| =

2 2

1 2+ .u u

1

2

u

u

a |u| =

b |v| =

c |

w| =

2 23 4 9 16 9 16 25

( ) +( ) = = = =15 5 25 25 25 25

2 23 + 0 = 9 = 3

2 22 +(3) = 4+9= 13Solution

EXAMPLE 13 Find the length of each vector.

a

u = (2, 3) b

v = (3, 0) c

w =3 4

( , )5 5

u = (3 1, 6 2) = (2, 4)

|u| =

2 22 + 4 = 4+16 = 20

Solution

EXAMPLE 14 Find the length of the vectoru with initial point (1, 2) and terminal point (3, 6).


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|

AB| =

(1 a)

2

+ 4 = 20(1 a)2 = 16

1 a = 4 or 1 a = 4

a = 3 or a = 5

2 2 2(1 ) +(4 2) = (1 ) + 4 =2 5a aSolution

5. Equal Vectors

Definition equalvectors

Two vectors are equal if and only if their corresponding scalar components are equal.

In other words, the vectorsu = (u1, u2) and

v = (v1,v2) are equal if and only if u1 =v1 and

u2 =v2.

u =

v , so a + b = 1 and 3 = b a.Solution

EXAMPLE 16u and

v are equal vectors with

u = (a + b, 3),

v = (1, b a). Find a and b.

Let the terminal point ofvbe (x, y), so

(x 2, y 4) = (3, 7)

x 2 = 3 and y 4 = 7 by the equality of vectors.

So x = 5 and y = 11.

Solution

EXAMPLE 17 The vectorv = (3, 7) has initial point (2, 4). What is its terminal point?


1 Find the position vector of

ML with endpointsM(3, 2) and L(2, 3).

2 Find the length of each vector.

a

u = (3, 1) b v = (0, 3) c AB with endpointsA(2, 1) and B(5, 4)

A n s w e r s

1

ML = (1, 1) 2 a 10 b 3 c 32

a + b = 1

b a = 3, so b = 1 and a = 2.

EXAMPLE 15 Find the possible values of a given |

AB| = 25 and the endpointsA(a, 2) and B(1, 4).


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EXAMPLE 22 Given

w = (1, 3), K(3, 2) and P(1, 4), find

w +

KP and

w

KP.

2. Subtraction of Vectors

Ifu = (u1, u2) and

v = (v1,v2) then

u

v = (u1 v1, u2 v2).

The figure opposite shows how the analytic definition

of vector subtraction corresponds to the geometric

one.

u

v = (2 3, 4 2) = (1, 2)Solution

EXAMPLE 21 Subtractv = (3, 2) from

u = (2, 4).

a. Properties of Vector Subtraction

Letu = (u1, u2),

v = (v1, v2), and

w = (w1, w2) be vectors in a plane. Then the following

properties hold.

1 The difference of any two vectors in a plane is a vector.

2

u

v

v

u. Therefore, vector subtraction is not commutative.3

u (

v

w ) (

u

v )

w . Therefore, vector subtraction is not associative.

4

u

0

0

u. Therefore there is no identity element for subtraction.

a. Properties of Vector Addition

Letu = (u1, u2),

v = (v1,v2), and

w = (w1,w2) be vectors in a plane. Then the following

properties hold.

1 The sum of any two vectors in a plane is a vector. (closure property)

2

u +

v =

v +

u (commutative property)

3

u + (

v +

w ) = (

u +

v ) +

w (associative property)

4

0 is the identity element:u +

0 = (u1 + 0, u2 + 0) = (u1, u2) =u.

5 u is additive inverse of

u:

u + (u) = (u1 u1, u2 u2).


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Geometry 9

3. Multiplication of a Vector by a Scalar

Letv = (v1,v2) and c , then c

v = (cv1, cv2).

2u = (2(3), (2)2) = (6, 4)

3v = (32, 3(1)) = (6, 3)

3u + 2

v = 3(3, 2) + 2(2, 1) = (9, 6) + (4, 2) = (5, 4)

Solution

EXAMPLE24

u = (3, 2) and

v = (2, 1) are given. Find 2

u, 3

v , and 3

u + 2

v .

a. Properties of the Multiplication of a Vector by a Scalar

Let

u = (u1, u2),

v = (v1,v2) and c, d . Then the following properties hold:1 c(

u +

v) = c

u + c

v. Look at the proof:

c(u +

v ) = c(u1 +v1, u2 +v2)

= (cu1 + cv1, cu2 + cv2)

= c(u1, u2) + c(v1,v2)

= cu + c

v.

2 (c + d)u = c

u + d

u

3 (cd)u = c(d

u ) = d(c

u )

4 1u =

u

5 0u =

0

6 c0 =

0

7 |cu| = |c||

u|.

KP = (1 3, 4 2) = (4, 2)w +

KP = (1, 3) + (4, 2) = (1 4, 3 + 2) = (3, 1)w

KP = (1, 3) (4, 2) = (1 + 4, 3 2) = (5, 5)

Solution

MN +

NK =

MK

|

MN +

NK| = |

MK| = 13

(m 1)2 + 25 = 169

(m 1)2 = 144

m 1 = 12 orm 1 = 12

m = 13 orm = 11

2 2( 1) +( 1 4) =13m

Solution

EXAMPLE 23 M(1, 4), N(3, 2m), K(m, 1), and |

MN +

NK| = 13 are given. Findm.


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3


4. Standard Base Vectors

a u 3v = (3, 1) 3(2, 2) = (3, 1) + (6, 6) = (3 6, 1 + 6) = (9, 7)

b 3u + 2

v = 3(3, 1) + 2(2, 2) = (9, 3) + (4, 4) = (9 + 4, 3 4) = (5, 1)

c 4u +

v = 4(3, 1) + (2, 2) = (12, 4) + (2, 2) = (12 + 2, 4 2) = (10, 2)

Solution

EXAMPLE 26 Find each vector, givenu = (3, 1) and

v = (2, 2).

a

u 3

v b 3

u + 2

v c 4

u +

v

2u 3

v = 2(2, 3) 3(1, 1) = (4, 6) (3, 3) = (4 3, 6 + 3) = (7, 9)

Therefore, |2

u 3

v | = 2 2

( 7) +9 = 49+81= 130.

Solution

EXAMPLE 25 Find |2u 3

v| given

u = (2, 3) and

v = (1, 1).


1 Find |w| given

u = (3, 2),

v = (1, 4),

w = (a, b) and

v

w = 3

u.

2 Findx given 2

x +

y = (1, 2), and

x

y = (4, 4).

A n s w e r s

1 |w| = 102 2

x = (1, 2)

u

|u|

1

|u|

= u 21

2 2 2 2

1 2 1 2

= , .+ +

uu

u u u u

u

|

u|

22 2 2 22

2 2 1 21 1

2 2 2 2 2 22 2 2 2

1 2 1 2 1 21 2 1 2

+= + = + = =1.

+ + ++ +

u u u uu u

u u u u u uu u u u

So

We sometimes use to find the direction ofu.

u

|u|

There are two important unit vectors,i and

j, defined as

i = (1, 0) and

j = (0, 1).

These vectors are special because we can use them to express any vector.

We call these vectors standardbasevectors.

Definition unitvector

A vector of length 1 is called a unitvector.

For any non-zero vectoru = (u1, u2), is a unit vector because and

1

|u|

2 21 2

1=

+u u

u

|u|

For example, the vectorw = is a unit vector.

3 4( , )5 5


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32

Geometry 9

b The properties of addition and scalar multiplication of vectors show that we can

manipulate vectors in the same way we manipulate algebraic expressions. Therefore,

2u + 3

v = 2(3

i + 2

j ) + 3(

i + 6

j )

= (6i + 4

j ) + (3

i + 18

j )

= (6 3)i + (4 + 18)

j

= 3i + 22

j.

a

u = 5

i + (8)

j = 5

i 8

j.Solution

EXAMPLE 27 a Write the vectoru = (5, 8) in terms of

i and

j.

b Ifu = 3

i + 2

j and

v =

i + 6

j , find 2

u + 3

v.

For example, let us express the vectorv = (v1,v2) in terms of

i and

j:

v = (v1,v2) =v1

i +v2

j.

In this expression,

v1

i is the horizontal component,v2

j is the vertical component,

v1 is the horizontal scalar component, and

v2 is the vertical scalar component.

We can prove the proposition above by using algebraic operations on vectors and the

properties of real numbers:v = (v1,v2) = (1v1 + 0, 0 + 1v2) = (1v1, 0) + (0, 1v2)

=v1(1, 0) +v2(0, 1)=v1

i +v2

j.

Letv be a vector in the plane with its initial point at the

origin. Let be the positive angle between the positive

x-axis andv (see the figure). If we know the length and

direction ofv , then we can resolve the vector into

horizontal and vertical components in terms of :v has length |

v|, and

v = (v1,v2) =v1

i +v2

j.

Sov1 = |v | cos and v2 = |

v | sin.

EXAMPLE 28 a |v| = 4 cm and the angle between

v and the positive x-axis is 60. Find the horizontal and

vertical components ofv and express

v in terms of

i and

j.

b Find the angle between the vectoru = 3

i +

j and the positive x-axis.


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1 Expressv = (3, 2) in terms of standard base vectors.

2 Resolve the vector u with length 3 cm into its components if the angle between the

positive x-axis andu is 120.

3 Find the angle between the vectorv = 3

i 33

j and the positive x-axis.

A n s w e r s

1

v = 3

i + 2

j 2

ux = ,

uy = 3 = 240

3 3(0, )

2

3( , 0)

2

a We havev = (v1,v2), where the scalar components are given by v1 = 4cos60 = 2 and

v2 = 4sin60 = 23. Therefore, the horizontal component is 2i and the vertical

component is 23j . Therefore,

v = 2

i + 23

j.

b From the figure we see that

has the property that

tan(180 ) = .

Thus 180 = 30, and so = 150.

1 3=

33

Solution

, so they are parallel.2 1

= = 2112

Solution

EXAMPLE 29 Show thatu = (2, 1) and

v = (1, ) are parallel.

1

2

EXAMPLE 30 Find the relation between x and y givenA(3, 1), B(2, 3), C(5, 4), D(x, y), and

CD ||

AB.

We know from the geometrical analysis of vectors that two non-zero vectors are parallel if and

only if multiplying one of them by a suitable scalar equals the other, that is,

for any c 0,u 0, and

v 0,

u ||

v if and only if

u = c

v.

It follows that ifu = (u1, u2) and

v = (v1,v2), then (u1, u2) = (cv1, cv2).

So

u || v ifandonlyif 21

2 2

v v

C. PARALLEL VECTORS

S l ti


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34

Geometry 9


1 Show that

u = (2, 3) and

v = (4, 6) are parallel.

2 Find the value of k ifu = (1, k) and

v = (3, 6) are parallel.

A n s w e r s

1 check the scalar components 2 k = 2

AB = (2 3, 3 (1)) = (1, 4)

CD = (x 5, y + 4)

CD ||

AB so

4x 20 = y 4, so 4x + y 16 = 0.

5 +4= .

1 4

x y

Solution

Let us write E(x, y) and F(m,n).

BE =

EA

BE = (x + 3, y + 2)EA = (3 x, 2 y)

x + 3 = 3 x

2x = 0

x = 0

y + 2 = 2 y

2y = 0

y = 0

and

CF =

FA

CF = (m 2,n + 3)FA = (3 m, 2 n)

m 2 = 3 m

2m = 5

m =

n + 3 = 2 n

2n = 1

n =1

2

5

2

Solution

EXAMPLE 31 A triangleABC has verticesA(3, 2), B(3, 2), and C(2, 3). E and F are the midpoints ofsidesAB andAC respectively. Find the coordinates of E and F.

emember

If two parallel vectorsa

andb have at least one

point in common, thena and

b lie on the same

straight line (they are

collinear).

For instance, if

AB = k

BC for some

k \ {0}, then A, B,

and C are collinear.

Therefore the coordinates are E(0, 0) and5 1

( , ).2 2

F


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35

Analysis of Vekctors Analytically

EXERCISES 1.2

A. Basic Concepts of Vectors in the

Analytic Plane

1. Plot the points A(1, 1), B(2, 1), C(3, 1), andD(3, 1) in the plane.

3. Find the length of

MN givenM(1, 4) and N(2, 1).

2. Sketch the position vector of the vector with the

given endpoints.a b c

EF

E(0, 3)

F(4, 2)

CD

C(1, 5)

D(0, 2)

AB

A(2, 3)

B(4, 1)

5. Describe the vector with initial point P and

terminal point Q.

4. The figure shows the

vectorsu and

v. Sketch

the following vectors.

a 2v b

u

c

u +

v d

u 2

v

e 2u +

v

c P(3, 2), Q(8, 9) d P(1, 3), Q(1, 0)

a b

Project:

Use The Geometers Sketchpad,

Cabri Geometry, or Javascript

sketchpad to sketch the vectors

2u,

v,

u +

v,

u 3

v and

PQ, QP if P(3, 4) and Q(4, 3).

We can use computer applications such as TheGeometers Sketchpad, Cabri Geometry, or

Javascript sketchpad to sketch vectors and solveproblems. We can use an application to aproblem, and then change certain values to seetheir effect. We can also use a computer

application to add and subtract vectors, and tomultiply a vector by a scalar.

The screen opposite shows a simple problem illustrated usingThe Geometers Sketchpad. Suppose a current flows at acertain velocity w downstream. A boat moves at a constantspeed v. Which direction the boat take in order to reach theother side of the river in the shortest possible time?

The Geometers Sketchpad calculates the time as we move thepoint H on the screen to set the direction of the boat. Using theSketchad we can move H to find the shortest possible time inthe problem.

6 Find the vectorsa and

b if 2

a 3

b (4 2) and

11 Given

AB 5i + 6

j and B( 4 8) find the


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36

Geometry 9

6. Find the vectors a and b if 2a 3b = (4, 2) anda +

b = (2, 1).

B. Vector Operations

7. Find u + v, u v, 2u, 2u v, u + 3v, and

4u + 3

v for the given vectors

u and

v.

a

u = (2, 6),

v = (1, 3)

b

u = (2, 3),

v = (8, 2)

c

u = (1, 0),

v = (0, 2)

d

u =

i

v =

j

e

u = 3

+

j

v =

i j

f

u = 7

i + 5

j

v =

j

i

8. Find |u|, |

v|, |2

v|, |

v|, |

u +

v|, and |

u

v|

for the given vectorsu and

v.

a

u = 3i + j v = i + 2j

b

u = 2

i

j

v =

i

j

c

u = (2, 3),

v = (0, 1)

d

u = (3, 4)

v = (2, 5)

1

3

9. Find the horizontal and vertical scalar components

of the vector with the given length and angle with

the positive x-axis. Express the vector in terms of

standard base vectors.

a |v | = 20, = 30

b |v | = 30, = 120

c |v | = 1, = 225

d |u | = 80, = 135

e |v | = 4, = 10

f |u | = 3, = 300

10. Given

AB = 7i + 2

j and B(3, 11), find the

coordinates of pointA.

11. Given AB = 5 i + 6j and B(4, 8), find the

coordinates of pointA.

12.u = 3

i + 4

j and

v = 4

i +

j are given. Which

vector is the longest?

13. Givenu = 3

i + 4

j, calculate |

u

i|.

1

2

14. Show thatu = (a, b) and

v = (2a, 2b) are

parallel.

16. In a triangle ABC, the vertices are A(2, 3),

B(0, 1) and C(4, 1). Points D(1, 2) and E(1, 2)

are on the sidesAB andAC respectively.

Show that

DE ||

BC.

17. In a triangle ABC, the vertices are A(1, 3),

B(2, 1), and C(3, 2). E( , 2) is on the sideAB.

Find the coordinates of F if F is on AC and

EF ||

BC.

3

2

15.u ||

v,u = (1, k 3), and

v = (k, k 4) are given.

Find the value of k.

C. Parallel Vectors


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38

Geometry 9

We have seen how to add and subtract vectors, and how to multiply vectors by a scalar.In this section we will introduce another operation on vectors, called the dotproduct.

The dot product is sometimes called the scalar product because the resulting product is a

number and not a vector. It has applications in mathematics, as well as in engineering and

physics.

A. DOT PRODUCT

Definitiondotproduct scalarproduct,orEuclideaninnerproduct)

Letu = (u1, u2) and

v = (v1,v2) be two vectors in the plane.

The dotproduct ofu and

v, denoted by

u

v, is defined by

= 11 + 22.

Thus, to find the dot product of two vectors we multiply the corresponding scalar components

and then add them together.

a.

u

v = 25 + 34 = 10 + 12 = 2

b.

u

v = 15 + 11 = 5 + 1 = 4

Solution

EXAMPLE 32 Find the dot product of the given vectors.a.

u = (2, 3) and

v = (5, 4) b.

u =

i +

j and

v = 5

i +

j

Proof1.

u

v = u1v1 + u2v2 =v1u1 +v2u2 =

v

u , by the commutative property of real numbers.

2.

u(

v +

w ) = (u1, u2)(v1 +w1, v2 +w2) = u1(v1 +w1) + u2(v2 +w2)

= u1v1 + u1w1 + u2v2 + u2w2

= u1v1 + u2v2 + u1w1 + u2w2

=u

v +

u

w

1. Properties of the Dot ProductThe definition of the dot product gives us the following properties.

1.

u

v =

v

u (commutative property)

2.

u(

v +

w ) =

u

v +

u

w (associative property)

3. c(u

v ) = (c

u)

v

4.

u

u = |

u|2

5.

u

v 0, and

u

u = 0 if and only if

u =

0.

3. c(u

v ) = c(u1v1 + u2v2) = cu1v1 + cu2v2 = (c

u )

v


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39

The Dot Product of Two Vectors

u

u = 22 + (2) (2) = 4 + 4 = 8. Therefore, |

u| = 8.Solution

EXAMPLE 33 Find the length of the vectoru = (2, 2) by using the dot product.


1. Find the dot product of

u = 3

j and

v =

i +

j.2. Find the quantity (

u +

v)(

u 3

v ) given

u =

j ,

v =

i.

A n s w e r s

1. 3 2. 2

Definition anglebetweentwovectors

Letu =

OE andv =

OFbe two non-zero vectors. The angle EOF

is called the angle between and . We use to represent the

smaller angle betweenu and

v when their initial points coincide.

1. Angle Between Two Vectors

Proof The proof is a nice application of the law of cosines.

As we know, |v

u|2 = (

v

u)(

v

u) =

v 2 +

u 2 2

v

u (1).

Applying the cosine law to triangle EOF in the figure,

|v

u|2 = |

u|2 + |

v|2 2|

u||

v| cos (2) (0 < < )

(v

u)(

v +

u) = |

u|2 + |

v|2 2|

u|2|

v|2cos

Theorem

Let be the angle measure between two non-zero vectorsu and

v. Then

u

v = |

u||

v|cos.

dotproducttheorem

B. ANGLE BETWEEN TWO VECTORS

cosinelaw:

a2 = b2 + c2 2bccos A

( ) ( 1 1 2 2) 1 1 2 2 ( )

4.

u

u = u1u1 + u2u2 = u

2

1 + u2

2 = |u|2, so

5. This proof is left as an exercise for you.

= |u|.u u

By (1) and (2),


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4

Geometry 9

EXAMPLE 34 Givenu = (2, 2),

v = (5, 8) and

w = (4, 3), find the following.

a.

u

v b. (

u

v )

w c.

u (2

v ) d. |

w|2

a. By the definition of the dot product we haveu

v = 2 5 + (2)8 = 6.

b. Using the result from (a) we have (u

v )

w = 6

w = 6( 4, 3) = (24, 18).

c. By property 3 of the dot product we have u (2v) = 2 (u v) = 2(6) = 12.

d. By property 4 of the dot product we have |

w|2 =

w

w = (4)2 + 32 = 25.

Solution

v 2 2

v

u +

u2 = |

u|2 + |

v|2 2|

u||

v|cos

2 v

u = 2|

u||

v|cos

Therefore,

u

v = |

u||

v|cos

.

One of the most important uses of the dot product is to find the angle between two vectors if

the scalar components of the vectors are given. We simply calculate the angle by solving the

equation given by the dot product theorem for cos . Let us state this important result clearly.

If is the angle measure between two non-zero vectorsu and

v then .

u

v

cos = ---------------------|

u||

v|

FINDING THE ANGLE BETWEEN TWO NON-ZERO VECTORS

a.

u

v = 2 5 + 5 2 = 20

|u| =

|v| =

by the formula, and so

b.

u

v = 1 2 + 2 22 = 2 + 4 = 6

|u| =

|v| =

Therefore, = 0.

6 6

cos = = =13 12 36

4+8 = 12

21 +2 = 3

20

cos = .29

20

cos =29 29

2 25 + 2 = 29

2 22 + 5 = 29

Solution

EXAMPLE 35 a. Find the cosine of the angle between the vectorsu = (2, 5) and

v = (5, 2).

b. Find the angle between the vectorsu = (1, 2) and v = (2, 22).



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Definition perpendicular orthogonal)vectors

Two non-zero vectorsu and

v are perpendicular if and only if the

angle measure between them is 90.

By the dot product thorem,u

v = |

u||

v|cos. We can conclude that for two non-zero

vectors

u and

v ,

u

v is zero if and only if equals 90.

2.Perpendicular and Parallel Vectors

Theorem

Two non-zero vectorsu and

v are perpendicular if and only if

u

v = 0.

u

v = 61 + (2)3 = 0So the vectors are perpendicular.

Solution

EXAMPLE 36 Are the vectorsu = (6, 2) and

v = (1, 3) perpendicular?

u

v = 32 + 5(6) = 6 30 = 24. So

u and

v are not perpendicular.Solution

EXAMPLE 37 Are the vectorsu = (3, 5) and

v = (2, 6) perpendicular?

1. Find the dot product ofu = (0, 3) and

v = (1, 2).

2. Find the length ofu = (1, 3) using the dot product.

3. Find (u

v )

w given

u = (0, 3),

v = (1, 2), and

w = (2, 1).

4. Find the angle betweenu = (2, 1) and

v = (1, 3).

A n s w e r s

1. 6 2. 10 3. (12, 6) 4. = 45

From the definition of parallel vectors we know thatu ||

v if and only if

u = k

v . We can

write,

|u| = |k||

v| (1)

u

v = k

v

v =k|

v|2 (2)

u

v = |

u||

v|cos (3).

From (1), (2) and (3),


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42

Geometry 9

|k||v||

v|cos = k|

v|2

cos =

cos = 1 or cos = 1 = 0 or = 180.

From these results we can conclude that two vectors are parallel if and only if the angle

measure between them is 0 or 180.

| |

k

k

Con c l u s i o n

Let be the angle measure between nonzero vectors

u and

v.

Then

|| ifandonlyif

=|

|

|

|or

=|

|

|

|.

Letv = (v1,v2) be perpendicular to

u.

u

v = 4v1 + 2v2 = 0

v2 = 2v1

Ifv1 = t R,v2 = 2t, thenv = (t, 2t).

Let t = 1 or 3:v = (1, 2) and

v = (3, 6) are both perpendicular to

u.

Solution

EXAMPLE 38 Determine two vectors in the plane which are perpendicular tou = (4, 2).

SoAD

BC and

AD

BC = 0.

Solution

EXAMPLE 39 In an equilateral triangleABC, D is the midpoint of BC. Find

AD

BC.

In an equilateral triangle, the median is also the

altitude, as show in the diagram.

EXAMPLE 40 In a squareABCD, E is the midpoint of side BC and |BC| = 4 cm. Find

AE

AB.

AB =a

b

Solution


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43


A(ABC) =

AH = (x0 2, y0 3)

BC = (3, 1)

AH

BC = 3(x0 2) + (y0 3) = 0

3x0 + y0 6 3 = 0

3x0 + y0 = 9

| | | |2

AH BC

AH = (0.4, 1.2)

|

AH| = 1

0.16+1.44 = 1.6 = 410

BC = (3, 1)

|

BC| = 9+1 = 10

2x0 x0y0 = 3y0 3 x0y0 + x0

x0 = 3y0 3

x0 3y0 = 3

3x0 + y0 = 9 (1)

x0

3y0

= 3 (2)

9x0 + 3y0 = 27

x0 3y0 = 3

10x0 = 24

x0 = 2.4 and y0 = 1.8.

BH = k

HC

BH = (x0, y0 1)

HC = (3 x0, 2 y0)

Solution

0 0

0 0

1=

3 2

x y

x y

A(ABC) = = 1 4

10 22 10

AE =a +

AE

AB =

a(

a + ) =

a 2 + =

a 2

Sincea

b,

a 2 = |

a|2 = 42 = 16.

2

a b

2

b

2

b

EXAMPLE 41 Find the area of the triangle with vertices A(2, 3), B(0, 1), C(3, 2).

+

Let us multiply (1)by 3. Then,

Theorem triangleinequality

If

d

i h l h Thi i ll d h


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44

Geometry 9

If u and v are vectors in the plane, then |

+ | |

|+|

|. This property is called the

triangleinequality.

By the property of the dot product, |

u +

v|2

= (

u +

v )2

=

u2

+ 2

u

v +

v2

.2

u

v 2|

u||

v|

{|u|2 + |

v|2 + 2

u

v } {|

u|2 + |

v|2 + 2|

u||

v|}

|u +

v|2 (|

u| + |

v|)2, since both |

u| + |

v| and |

u +

v| are non-negative.

Therefore, |u +

v| |

u| + |

v|.

Proof

Theoremu and

v are perpendicular in the plane if and only if |

u +

v|2 = |

u|2 + |

v|2.

Proof |u +

v|2 = (

u +

v )2 =

u 2 + 2

u

v +

v 2 = |

u |2 + 2

u

v + |

v |2

= |u |2 + |

v |2, since

u

v = 0.

|2u 3

v|2 = (2

u 3

v)2 = 4

u2 + 9

v2 12

u

v

= 4|u|2 + 9|

v|2 12|

v||

v|cos

= 432 + 942 1234

= 36 + 144 72

= 108.

Now |2u 3v|2 = 108, and so|2u 3v| = 108 = 63.

1

2

Solution

EXAMPLE42 |

u|= 3, |

v|= 4, and the angle between

u and

v is 60. Find |2

u 3

v|.

Let us choose avariable point B(x, y):

AB = (x + 1, y 3).

AB and

n are perpendicular, so

ABn = 0.

3(x + 1) + 5(y 3) = 0

3x + 3 + 5y 15 = 0

3x + 5y 12 = 0 is the required equation.

Solution

EXAMPLE 43 Find an equation for the line passing throughA(1, 3) which is perpendicular ton = (3, 5).


Fi d th l f if

( 1) i di l t

(3 4)


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45


1. Find the value of a if u = (a, 1) is perpendicular to v = (3, 4).

2. In the right triangleABC,AC is the hypotenuse, BHAC, and H lies onAC.

Find (

HC +

CB)(

AB +

BH).

3. Find an equation for the line passing through P(3, 1) which is perpendicular ton = (3, 1).

A n s w e r s

1. 2. 0 3. 3x + y 8 = 0 4

3a


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46

Geometry 9


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47


ACROSS

1. A vector with length zero.

5. An undefined concept in geometry which has noshape or size.

7. Being the main or most important part.

10. Being parallel.

12. The coordinate representing the position of a pointalong a line perpendicular to the y-axis in a plane.

13.A thin straight stick with a point, sometimes used toshoot things.

15.A mathematical statement that establishes theequality of two expressions.

16.Any of the four areas into which a plane is dividedby the reference axes in a rectangular coordinatesystem, designated first, second, third, and fourth,counting counterclockwise from the area in which

both coordinates are positive.

17. One of the two horizontal or vertical vectors whosesum is equal to a given vector.

18. In a triangle, the formula cosA = (b2 + c2 a2)/2bc.

21. Vectors which have the same direction and length.

24. To arrange in sets of two.

25. The act or process of adding.

26. The direction 90 clockwise from north.

27. Either of two points marking the end of a linesegment.

DOWN

2. A single vector that is the equivalent of a set ofvectors.

3. The measure of heat of a body or environment.

4. The principal structural member of a ship.

6. The solution of an equation in which every variableis equal to zero.

7. An acute angle measured from due north or duesouth.

8. The size of a flat surface, calculated mathematically.

9. The property that states: if a, b A, then a b Afor an operation.

11.A point whose position is constant.

14. The direction 270 clockwise from due north anddirectly opposite east.

19.A picture such as a pie chart or bar graph, used toillustrate quantitative relationships.

20. To give the meaning of (a word or idea).

22.An undefined concept in geometry that describes aset of points along a path.

23.A relatively small, usually open craft used byfishermen.

EXERCISES 1.3


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48

Geometry 9

A. Dot Product

1. Find the dot product of each pair of vectors.

a.

u = (2, 1) v = (3, 6)

b.

u = 7

i + 24

j,

v = 3

i

j

c.

u = 3i + j, v = 3i j

d.

u =

i

j,

v = 2

i

e.

u = 19

j,

v = 16

j

2. Find the following quantities given

u =

i + 2

j,v = 3

i +

j , and

w = 4

i + 3

j.

a.

u

w +

u

w

b.

w(

u +

v )

c. (u +

w )(

u

w)

d. 2(u

v )

e. (

w

v )(

w +

u )f. (3

u 2

v )(

u + 2

v )

d.

u = 2

i 8

j ,

v = 12

i + 3

j

e.

u = 3

i 4

j ,

v = 8

i 6

j

f.

u = 4i , v = 3i

8. Find two perpendicular vectors to each given

vector.

a.

a = 3

i

j b.

b = 8

i 6

j c.

c =

i + 2

j

9. For what values of t areu = 4

i 5

j t and

v = 3

i 2

j perpendicular?

10. |u| = 5 and |

v| = 3 are given. For what values

of t areu + t

v and

u t

v perpendicular?

12. Show that the equality |

u +

v| = |

u

v| holdswhen

u and

v are orthogonal.

11. For what values of t doa = t

i +

j and

b =

i + t

j have angle measure 120 between

them?

7. Find the measure of the interior angles of the triangle

ABC with verticesA(1, 3), B(1, 2), and C(2, 2).

4. Evaluate the following quantities if the angle measure

betweenu and

v is 60 and |

u| = 4, |

v| = 3.

a.

u

v b. |

u +

v|2

c. |u

v|2 d. (3

u + 2

v ) (

u + 2

v )

5. Evaluate |u +

v| given |

u| = 13, |

v| = 5, and

|u

v| = 12.

3.u,

v and

w are unit vectors such that

u +

v +

w = 0. Find

u

v +

v

w +

w

u = 0.

13. Verify the equality

|u +

v|2 + |

u

v|2 = 2(|

u|2 + |

v|2).

B. Angle Between Two Vectors

6. Determine whether the given vectors are

perpendicular or not.

a.

u = (4, 6),

v = (3, 2)

b.

u = (5, 0),

v = (0, 4)

c.

u = 3

i ,

v =

j

14. In a rhombusABCD, one side measures 6 cm and

E, F are the midpoints of sides AD and DC

l Th l b d AD

Find the area of a triangle with vertices

A(2, 2), B(0, 2), and C(1, 4).

21.


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49


respectively. The angle measure between sidesAD

and DC is 120. Find the quantity

BE

AF.

15. In a rectangleABCD,

DC = 3

AD and point E is on

DC. Find the quantity

AE

BE given

DE = 2

EC

and |

AD| = 3 cm.

16. Show that the inequality |u +

v| > |

u

v| holds

when the angle measure betweenu and

v is less

than 90.

17. Find two unit vectors that make angles of 45

with i +

j.

19. Write the equation of the line passing through

A(1, 1) which is perpendicular to

u = (3, 4).

18. Letu and

v be vectors and let be a scalar. Verify

the given properties.a.

u

v =

v

u

b. (u )

v = (

u

v ) =

u(

v )

Show that the diagonals of a rhombus are

perpendicular using vectors.

20.

Find the area of a rhombus with vertices A(2, 0),

B(3, 3), C(8, 0), and D(3, 3).

22.

Find the area of a rectangle with verticesA(3, 2),

B(9, 2), C(9, 5), and D(3, 5).

23.

Show that the altitudes of an acute-angled triangle

are concurrent using vectors.

25.

Find the distance of P(x0, y0) from the line

ax + by + c = 0 using vectors.

26.

Find the area of a parallelogram with vertices

A(2, 1), B(6, 0), C(8, 3), and D(4, 2).

27.

For any vectors

u,

v, and

w prove that(

u

w )

v

u(

v

w) is perpendicular to

w.

24.

CHAPTER SUMMARY

A li g t ith di ti i ll d di t d li


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5

Geometry 9

A line segment with direction is called a directed line

segment.

A vector in the plane is a directed line segment.

AdditionofTwoVectorsGeometrically

For two vectors

PO and

QR in a plane,

PQ +

QR is the

sum ofPQ and

QR. There are two ways to add vectors

geometrically: the polygon method and the parallelogram

method.

ThePolygonMethod

In this method we draw the first vector. Then we place

the initial point of the second vector at the terminal point

of the first, the initial point of the third vector at theterminal point of the second, and so on until we place the

initial point of the nthvector at the terminal point of the

(n 1)th vector. The vector whose initial point is the

initial point of the first vector and whose terminal point

is the terminal point of the last vector is the resultant

(sum) vector.

TheParallelogramMethod

In this method we draw the first vector, and then draw

the second vector with its initial point at the initial pointof the first vector. We make a parallelogram by drawing

two additional sides, each passing through the terminal

point of one of the vectors and parallel to the other

vector. The sum is drawn along the diagonal from the

common initial point to the intersection of the two lines.

MultiplicationofaVectorbyaScalar

For a real number a and a vectoru:

If a > 0, then the vector au has the same direction

to u and the length |au| = a|u|.

If a < 0, then the vector au has the opposite

direction tou and the length |a

u| = |a||

u|.

If a = 0, then au =

0.

ParallelVectors

Leta and

bbe two vectors.

a and

b are parallel if and

only ifa = k

b where k 0.

Non-ParallelVectorIf

a is not parallel to

b, then h

a =k

b when h = k = 0.

ComponentsofaVectors

A vector in a plane has two components, called the

horizontal and vertical components of the vector.

Let |a| = (a1, a2) be a vector, then |

a| = 2 2

1 1+ .a a

EqualVectors

Two vectors are equal if and only if their correspondingcomponent vectors are equal.

In other words, for a = (a1, a2) and b = (b1, b2), thena =

b if and only if a1 = b1 and a2 = b2.

StandardBaseVectorsi = (1, 0) and

j = (0, 1) are called the standard base

vectors.

LinearCombinationofVectors

Letu1,

u2, ... ,

ukbe vectors in a plane and let c1, c2, ... ck

be scalars.v = c1

u1 + c2

u2 + ... + ck.

uk is called a

linear combination of vectors.

DotProduct

The dot product ofu = (u1, u2) and let

v = (v1,v2) is the

scalar quantityu

v = u1v1 + u2v2.

AngleBetweenTwoVectors

Let be the angle between two non-zero vectorsu and

v.

1. u ||

v if and only if

u

v = |

u||

v| or

u

v = |

u||

v| because

= 0 or

= 180.2.

u and

v are perpendicular if and only if

u

v = 0.

1. Can we use directed line segments in traffic? Give anexample.

2. What is the difference between a vector and a scalar?

3. Is it possible to add a hundred vectors in a plane using thepolygon method or the parallelogram method?

4. How do you change the direction of a vector using a realnumber?

5. a. Draw a diagram to show how to add two vectors.

b. Draw a diagram to show how to subtract two vectors.

6. Can you equalize two non-parallel vectors usingmultiplication by two real numbers?

7. In a plane, how many components do you need to resolvea vector?

8. How many standard base vectors are there in the plane?

9. What is the difference between the inner product and dotproduct of two vectors?

10.How do you use the dot product to find the angle measurebetween two vectors?

11.How do you use the dot product to determine whether twovectors are parallel or not?

12.How do you use the dot product to determine whether twovectors are perpendicular or not?

Concept Check

CHAPTER REVIEW TEST 1A

1 I t i l ABC G i th t id D ib


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5

Chapter Review Test 1A

5. (1, 4) +v = (4, 5) is given. Find

v.

A) (3, 9) B) (5, 1) C) (5, 1)

D) (1, 5) E) (9, 3)

8. Find the unit vector with opposite direction to

AB givenA(6, 3), and B(2, 6).

A) B) C)

D) E) 4 3

( , )5 5

4 3

( , )7 7

4 3

( , )7 7

4 3

( , )5 5

4 3

( , )5 5

3. In the figure,

|AB| = |BC| = |CD|

and the vectorsa,

b,

c,

d

are given. Which one of

the following is the linearcombination of

c in terms of

a and

d?

A) B) C)

D) E)

a + 2

d

-------------------2

a + 2

d

-------------------3

2a +

d

-------------------3

2a +

d

-------------------2

a +

d

----------------2

7.u = 2

i + 3

j and

v =

i 2

j are given.

Find 3u 2

v.

A) 4i + 5

j B) 8

i + 5

j

C) 8i + 13

j D) 4

i + 13

j

E) 4i + 13

j

1. In triangle ABC, G is the centroid. Describe

AG +

GC +

CA.

A)0 B)

AG C)

BG D) 2

CA E)

AC

10. Describe the unit vectora in the

figure.

A) i + j B) i + j

C)

i j D)

i +

j

E)

i3

2

3

2

1

2

3

2

1

2

12

3

2

12

3

2

4.

AD =

AB and

BE =

BC are given. Express

DE in terms of

AB and

BC.

A)

AB +

BC B)

AB +

BC

C)

AB

BC D)

AB

BC

E)

BC

AB

3

4

1

2

1

2

1

4

1

2

3

4

1

2

1

4

1

2

3

4

12

14

2. In the figure, K, L,M, N, P, R are the midpoints of

the sidesABCDEF

respectively. Describe

KB+

LC+

MD+

NE+

PF.

A)

AR B)

RA C)

FK D)

FA E)

PA

6. Find the coordinates of B if

AB = (7, 3) and

A(2, 1).

A) (9, 4) B) (9, 2) C) (9, 4)

D) (9, 4) E) (9, 2)

9.u =

i 5

j and

v = 2

i + 3

j are given. Find

3u + 2

v.

A) 3i 2

j B) 5

i

j C) 7

i 9

j

D) 2i + 3

j E) 8

i

j

16. ABCD is the rectangle in

the figure.

11. For how many values ofm > 0 area = (2, 1 m)

andb = (m + 1, 4) parallel?


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Geometry 9

15. Which one of the following is perpendicular tov = (3, 4)?

A) B) (4, 3) C)

D) E) 4 3

( , )5 5

3 4

( , )5 5

3 4( , )5 3

4 3

( , )5 5

FindABAC if

|AB| = 4 and

|BC| = 3.

A) 9 B) 12 C) 15 D) 16 E) 20

18. G is the centroid of a triangle with vertices

A(1, 1). B(4, 2), and C(0, 6). Find the scalar

product of

AB

BG.

A) 8 B) 14 C) 0 D) 1 E) 12

20. ABC is a triangle with

|AB| = 4,|AC| = 5,

|BC| = 6. Find

AC (

AB +

BC).

A) 20 B) 24 C) 25 D) 30 E) 50

17. In the figure, D and E

are the midpoints of

the sides of triangle

ABC.

Find (

AD

AE)

BC.

A) 100 B) 150 C) 0 D) 50 E) 100

19. A triangleABC has verticesA(1, 1), B(4, 2), and

C(0, 6). Find the scalar component of

AB alongBC.

A) B) C) D) E) 3

2

3

4

3

4

4

2

4

2

12. Which one of the following is true for vectorsa = (3, 6),

b = (6, 7) and

c = (9, 13)?

A)a = 4

b + 3

c B)

a = 3

b + 4

c

C)

a = 4

b 3

c D)

a = 3

b 4

c

E)a =

b +

c

A) 1 B) 2 C) 3 D) 4 E) 5

14. A(2, 5), B(1, 3), C(m, 6), and

AB

BC are given.

Findm.

A) 3 B) 2 C) 1 D) 2 E) 3

13. Find the measure of the angle between the unit

vectors

a and

b in degrees if their dot product is

A) 45 B) 60 C) 120 D) 135 E) 150

1.2

CHAPTER REVIEW TEST 1B

1 |

| |

|

6 h h l A C A [AC]


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53

Chapter Review Test 1A

1. |a| = 3 |b| = 5, and the angle between

a and

b is 60. Find |

a +

b|.

A) 7 B) 8 C) 9 D) 10 E) 11

7. What is the angle betweena = (1, 3) and

b = (4, 43) in degrees?

A) 60 B) 120 C) 135 D) 150 E) 180

8. A squareABCD has side

3 cm, and K, L are the

midpoints of sidesAB and

BC respectively. P and Q

trisectAC as shown in the

figure. FindDP

PQ.

A) 0 B) 1 C) 1 D) 2 E) 3

9.a,

b, and

c are three vectors such that

b =

a 2

c,

b

c, and |

a| = 4|

c|. Find the angle between

a and c in degrees.

A) 30 B) 45 C) 60 D) 75 E) 90

10. Which one of the following is false for two vectorsa,

b and a scalar k ?

A) If a b, ab = 0.

B) Ifa ||

b,

b

a = 0.

C)a

b =

b

a

D) (ka + k

b) = k(

a +

b)

E) (kak

b) = k2

a

b

3. Given |a| = 12, |

b| = 5, and |

a

b| = 8, find

cos(a,

b).

A) B) C) D) E) 12

13

5

13

3

4

7

8

3

8

4. In an equilateral triangleABC, D [AC], |BC| = 6,

and |CD| = 2. Find

BC

BD.

A) 30 B) 24 C) 20 D) 18 E) 12

5. |a| = 7, |

b| = 10, and |

a +

b| =73 are given.

Find |a

b|.

A) 3 B) 8 C) 13 D) 15 E) 17

2. In the figure,

|AB| = |AC| = 4,

mA = 120, and

|AD| = |DC|.

Find

BC

BD.

A) 48 B) 36 C) 30 D) 24 E) 18

6. In the right triangleABC, mA = 90, D [AC],

and |AB| =k. Find

BA

BD.

A)k B)k2 C) D) E) 2k22

k2

2

k

11. In the figure, ABCD is a

parallelogram and

|BC| = 1,

16. In a triangle ABC,

AB = (4, 2a),

AC = (a, 4)

and the length of

BC is 10 cm. Find a possible

value of a


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Geometry 9

| |

|DC| = 2,

|DE| = |EC|. Find(

BC +

CE)(

AD +

DE).

A) 0 B) 1 C) 2 D) 3 E) 4

12. In an equilateral triangle ABC, D and E are the

midpoints of [AC] and [BC] respectively.

Find

DC (

BA +

AE).

A) |

AB|2 B) |

EB|2 C) |

DC|2

D) |

DC|2 E) |

EB|1

2

1

2

1

2

13. In an equilateral triangle ABC with side 1 cm,

what is

BC

CA?

A) B) C) D) 2 E) 3

2

3

2

1

2

1

2

14. For what values of k are the vectorsa = (12k, 9)

and b = (4, 3) linearly dependent?

A) 3 B) 2 C) 1 D) 0 E) 1

value of a.

A) 1 B) 2 C) 3 D) 4 E) 5

18.a +

b = (1, 3) and 2

a +

b = (4, 6) are given.

Find the measure of the angle betweena and

b.

A) 30 B) 45 C) 90 D) 135 E) 150

19. In a triangleABC,

AB = (2, 5) and

AC = (2, 2). Find the length of

BC.

A) 1 B) 2 C) 3 D) 4 E) 5

20.a = (12, 5), and

b = (3, 4) are given. Find

sin(a,

b).

A) B) C) D) E) 4

13

10

13

63

65

12

13

5

13

17. In a squareABCD

with side 2 cm,

|AE| = |ED| and

|DF| = |FC|.Find

EF (

EA +

AB).

A) 1 B) 2 C) 2 D) 1 E) 0

15. In a square ABCD, the

side is 4 cm,

[AB] [EP], [PD] [PC],and |AE| = |EB|.

Find

PE (

PC +

PD).

A) 16 B) 12 C) 10 D) 8 E) 6


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56

Geometry 9

A. THE CONCEPT OF CONGRUENCE

1. Congruent Figures and Polygons

The world around us is full of objects of various shapes and sizes. If we tried to compare some

of these objects we could put them in three groups:

objects which have a different shape and size,

objects which are the same shape but a different size, and

objects which are the same shape and size.

The tools in the picture at the right have different shape and size.

The pictures below show tools which have the same shape but different size. In geometry,

figures like this are called similarfigures. We will study similar figures in Chapter 3.

The pictures below show objects which are the same size and shape.

In this section, we will study figures which have this property.Factories often need toproduce many parts

with exactly the samesize and shape.

Definition congruentfigures

Figures that have the same size and shape are called congruentfigures. We say A is congruent

to B (or B is congruent toA) ifA and B are congruent figures.


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Congruence

The pictures at the bottom of the previous page show some examples of congruent objects.

The pictures below show two more examples. In these two examples there is only one piece

left to fit in the puzzle. Therefore, without checking anything, we can say that each piece and

its corresponding place are congruent.

Congruence in nature:the petals of this flower

are congruent.

Challenge

Remove five toothpicksto make five congruenttriangles.

In the figure below, ABC and DEF are congruent because their corresponding parts are

congruent. We can write this as follows:

A D AB DE

B E and BC EF

C F AC DF.

We can show this symbolically in a figure as follows:

Definition congruenttriangles

Two triangles are congruent if and only if their corresponding sides and angles are congruent.

We write C DEF to mean that ABC and DEF are congruent.

2. Congruent Triangles

EXAMPLE 1 Given that MNP STK, state the congruent angles and sides in the two triangles withoutdrawing them.


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Geometry 9

Solution M S MN ST

N T and NP TK

P K PM KS.

As we can see, the order of the vertices in congruent triangles is important when we are

considering corresponding elements. Any mistake in the ordering affects the correspondence

between the triangles.If two triangles are congruent then we can write this congruence in six

different ways. For instance, if ABC is congruent to DEF, the following statements are all

true:

ABC DEF

ACB DFE

BAC EDF

BCA EFD

CAB FDE

CBA FED.

A short history of the symbol:

GottfriedWilhelm

Leibniz

(1640-1716)

introduced for congruence

in an unpublished manuscript in

1679.

In 1777,

JohannFriedrich

Hseler

(1372-1797)

used (with the tilde reversed).

In 1824,

Carl randan

Mollweide

(1774-1825)

used the modern symbol for

congruence in Euclids Elements.

If two triangles are congruent then we can write this congruence in six different ways. For

instance, if ABC is congruent to DEF, the following statements are all true:

ABC DEF

ACB

DFE

BAC EDF

BCA EFD

CAB FDE

CBA FED.

EXAMPLE

2 Complete each statement, given that

PRS

KLM.a. PR _____ b. _____ K c. _____ SP

d. S _____ e. ML _____ f. L _____

Solution a. PR KL b. P K c. MK SP

d. S M e. ML SR f. L R

EXAMPLE 3 Decide whether or not the two triangles inthe figure are congruent and give a reason for

your answer.


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59

Congruence

y

Solution Let us calculate the missing angles:

m(C) = 60 (Triangle Angle-Sum Theorem in ABC)

m(M) = 30 (Triangle Angle-Sum Theorem in KMN)

Now we can write the congruence of corresponding parts:

AB KM (Given)

BC KN (BC = KN = 4)

AC MN (AC =MN = 8)

A M (m(A) =m(M) = 30)

B K (m(B) =m(K) = 90)

C N (m(C) =m(N) = 60)

Therefore, ABC MKNby the definition of congruent triangles.

EXAMPLE 4 ABC EFD is given with AB = 11 cm, BC = 10 cm and EF + ED = 19 cm. Find theperimeter of EFD.

Solution Since ABC EFD,AB = EF, BC = FD and

AC = ED by the definition of congruence.

So by substituting the given values we get

11 = EF, 10 = FD andAC = ED.

Since we are given that EF + ED = 19 cm,

we have 11 + ED = 19 cm; ED = 8 cm.

So P(EFD) =EF + ED + FD = 11 + 8 + 10 = 29 cm.


1. KLM XYZ is given. State the corresponding congruent angles and sides of the

triangles.


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Geometry 9

2. State the congruence JKM SLX in six different ways.

3. Triangles KLM and DEF are congruent. P(KLM) = 46 cm, the shortest side of KLM

measures 14 cm, and the longest side of the DEF measures 17 cm. Find the lengths of

all the sides of one of the triangles.

4. Triangles DEF and KLM are congruent. If DE = 12.5 cm, EF = 14.4 cm and the perimeter

of the triangle KLM is 34.6 cm, find the length of the side DF.

5. Two line segments KL andABbisect each other at a point T. IfAL = 7 and the lengths of

the segments KL andAB are 22 and 18 respectively, find the pe

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9 Geometry